4.
若\( \omega \)為1的立方虛根之一，則\( (1-\omega+\omega^2)(1-\omega^2+\omega^4)(1-\omega^4+\omega^8)(1-\omega^8+\omega^16)= \)？
(A)8 (B)10 (C)12 (D)14 (E)16
[解答]
\( (1-\omega+\omega^2)(1-\omega^2+\omega)(1-\omega+\omega^2)(1-\omega^2+\omega) \)
\( = [\ (1-\omega+\omega^2)(1-\omega^2+\omega) ]\ ^2= [\ (-2 \omega)(-2 \omega ^2)]\ ^2=16 \)
16.
在等比數列\( \langle a_{n} \rangle \)中，\( a_{1}=1 \)，\( a_{4}=2-\sqrt{5} \)，\( a_{n+2}=a_{n+1}+a_{n} \)，\( n \ge 1 \)。則\( \langle a_{n} \rangle \)的公比=？(A)\( \frac{1-\sqrt{5}}{2} \) (B)\( \frac{1+\sqrt{5}}{2} \) (C)\( \frac{1-\sqrt{3}}{2} \) (D)\( \frac{1+\sqrt{3}}{2} \) (E)\( \frac{2}{3} \)
[提示]
\( \displaystyle \frac{a_{n+2}}{a_{n+1}}=1+\frac{a_n}{a_{n+1}} \)，\( \displaystyle r=1+\frac{1}{r} \)
取負的r
40.
如右圖，三個兩兩外切的圓，也都與直線相切，最大圓半徑為144，中圓的半徑為36，求最小圓的半徑為何？(A)4 (B)12 (C)16 (D)18。
[提示]
\( \displaystyle \frac{1}{\sqrt{r}}=\frac{1}{\sqrt{144}}+\frac{1}{\sqrt{36}} \)
2009.10.14補充
Two circles with radii a and b respectively touch each other externally. Let c be the radius of a circle that touches these two circles as well as a common tangent to the two circles. Prove that \( \displaystyle \frac{1}{\sqrt{c}}=\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}} \)
http://www.mathlinks.ro/viewtopic.php?p=1597647
109.6.6補充
三個兩兩外切的圓，也都與直線相切，最大圓半徑為100，中圓的半徑為25，求最小圓的半徑為何？(A)\(\displaystyle \frac{100}{9}\) (B)\(\displaystyle \frac{10}{3}\) (C)\(\displaystyle \frac{36}{5}\) (D)\(\displaystyle \frac{18}{5}\)
(109全國高中職聯招，
https://math.pro/db/thread-3342-1-1.html)
44.
設z，c皆為複數，|z|=1，\( \overline{c} z≠1 \)，z≠c，則\( \displaystyle \Bigg\vert\ \frac{z-c}{1-\overline{c} z} \Bigg\vert\ \)之值為？
(A)0 (B)1 (C)∞ (D)|c| (E)以上皆非
[解答]
|z|=1，\( z \cdot \overline{z}=1 \)
\( \displaystyle \frac{z-c}{1-\overline{c} z} \cdot \frac{\overline{z}-\overline{c}}{1-c \overline{z}}=\frac{z \cdot \overline{z}-z \cdot \overline{c}-c \cdot \overline{z}+c \cdot \overline{c}}{1-c \cdot \overline{z}-\overline{c} \cdot z+c \cdot \overline{c} \cdot z \cdot \overline{z}}=1 \)
50.
設a,b,c均為整數，1≦a,b,c≦9，已知a,b,c成等差數列，且\( 0. \overline{a}+0. \overline{4b}=1. \overline{2c} \)，則序組(a,b,c)=
(A)(7,5,3) (B)(7,6,5) (C)(8,6,4) (D)(8,7,6) (E)(6,7,8)
[解答]
\( \displaystyle \frac{a}{9}+\frac{40+b}{99}=1+\frac{20+c}{99} \)
用c=2b-a來換得\( 12a-b=79 \)只有a=7符合
